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Question Number 88310 by ajfour last updated on 09/Apr/20
Commented by ajfour last updated on 09/Apr/20
Q.88272 (revisited)
Q.88272(revisited)
Answered by ajfour last updated on 10/Apr/20
h^2 +k^2 =r^2 y^2 =b^2 (1−(((x−a)^2 )/a^2 )) [ellipse] x^2 +y^2 −2hx−2ky=0 [circle] (x^2 +y^2 −2hx)^2 =4(r^2 −h^2 )y^2 ⇒ {x^2 +b^2 (1−(((x−a)^2 )/a^2 ))−2hx}^2 = 4(r^2 −h^2 )b^2 (1−(((x−a)^2 )/a^2 )) roots of above eq. are x_1 =0 , x_2 =2h, x_3 =x_4 =t first we eliminate x=0, then determine h(r). then eliminate t and eventually find r. .... let (x/a)=X, (h/a)=s, (b^2 /a^2 )=q, (t/a)=m, (r/a)=R . {X^2 +q(2X−X^2 )−2sX}^2 = 4(R^2 −s^2 )q(2X−X^2 ) ⇒ X{X+q(2−X)−2s}^2 =4q(R^2 −s^2 )(2−X) roots are now X=2s, m,m. lets put X=2s ⇒qs(1−s)=(R^2 −s^2 ) ⇒ R^2 =(1−q)s^2 +qs ...(I) And R^2 −s^2 =qs(1−s) ...(II) now generally (1−q)^2 X^3 +4(1−q)(q−s)X^2 +4(q−s)^2 X +4q(R^2 −s^2 )X−8q(R^2 −s^2 )=0 ⇒ (1−q)^2 X^3 +4(1−q)(q−s)X^2 +4{(q−s)^2 +q(R^2 −s^2 )}X −8q(R^2 −s^2 )=0 2m+2s=−((4(q−s))/((1−q))) ...(i) 2m^2 s=((8q(R^2 −s^2 ))/((1−q)^2 )) ...(ii) ⇒ s{((2(s−q))/((1−q)))−s}^2 =((4q(R^2 −s^2 ))/((1−q)^2 )) now using (II) ⇒ s{((2(s−q))/((1−q)))−s}^2 =((4q(qs)(1−s))/((1−q)^2 )) ⇒ {((2(s−q))/((1−q)))−s}^2 =((4q^2 (1−s))/((1−q)^2 )) ⇒{(1+q)s−2q}^2 =4q^2 (1−s) (1+q)^2 s^2 −4qs=0 and as s≠0 s=((4q)/((1+q)^2 )) Now using (I) ⇒ R=(√((1−q)s^2 +qs )) ⇒ R = (√(((16(1−q)q^2 )/((1+q)^4 ))+((4q^2 )/((1+q)^2 )))) or R=((2q(√((1−q)^2 +4)))/((1+q)^2 )) for example if a=4, b=3 q=(9/(16)), R=((18(√(1073)))/(625)) ≈ 0.9434 ...
h2+k2=r2y2=b2(1−(x−a)2a2)[ellipse]x2+y2−2hx−2ky=0[circle](x2+y2−2hx)2=4(r2−h2)y2⇒{x2+b2(1−(x−a)2a2)−2hx}2=4(r2−h2)b2(1−(x−a)2a2)rootsofaboveeq.arex1=0,x2=2h,x3=x4=tfirstweeliminatex=0,thendetermineh(r).theneliminatetandeventuallyfindr.….letxa=X,ha=s,b2a2=q,ta=m,ra=R.{X2+q(2X−X2)−2sX}2=4(R2−s2)q(2X−X2)⇒X{X+q(2−X)−2s}2=4q(R2−s2)(2−X)rootsarenowX=2s,m,m.letsputX=2s⇒qs(1−s)=(R2−s2)⇒R2=(1−q)s2+qs…(I)AndR2−s2=qs(1−s)…(II)nowgenerally(1−q)2X3+4(1−q)(q−s)X2+4(q−s)2X+4q(R2−s2)X−8q(R2−s2)=0⇒(1−q)2X3+4(1−q)(q−s)X2+4{(q−s)2+q(R2−s2)}X−8q(R2−s2)=02m+2s=−4(q−s)(1−q)…(i)2m2s=8q(R2−s2)(1−q)2…(ii)⇒s{2(s−q)(1−q)−s}2=4q(R2−s2)(1−q)2nowusing(II)⇒s{2(s−q)(1−q)−s}2=4q(qs)(1−s)(1−q)2⇒{2(s−q)(1−q)−s}2=4q2(1−s)(1−q)2⇒{(1+q)s−2q}2=4q2(1−s)(1+q)2s2−4qs=0andass≠0s=4q(1+q)2Nowusing(I)⇒R=(1−q)s2+qs⇒R=16(1−q)q2(1+q)4+4q2(1+q)2orR=2q(1−q)2+4(1+q)2forexampleifa=4,b=3q=916,R=181073625≈0.9434…
Commented by mr W last updated on 10/Apr/20
it′s perfectly solved!
it′sperfectlysolved!
Commented by ajfour last updated on 10/Apr/20
great, thanks for the confirmation, Sir, I′m too glad i cud solve it.
great,thanksfortheconfirmation,Sir,I′mtoogladicudsolveit.
Commented by mr W last updated on 09/Apr/20
that′s right sir.
that′srightsir.
Commented by ajfour last updated on 10/Apr/20
thanks to mjS Sir.
thankstomjSSir.
Commented by ajfour last updated on 10/Apr/20
Commented by MJS last updated on 10/Apr/20
I tried the same path I used with the parabola and also get R=((18(√(1073)))/(625)) for (a/b)=(4/3) but there should be a scaling factor, I calculated with a=1∧b=(3/4)
ItriedthesamepathIusedwiththeparabolaandalsogetR=181073625forab=43butthereshouldbeascalingfactor,Icalculatedwitha=1∧b=34
Commented by ajfour last updated on 10/Apr/20
so our answer should be right then..? required radius r=aR, Sir.
soouranswershouldberightthen..?requiredradiusr=aR,Sir.
Commented by mr W last updated on 10/Apr/20
i checked my result again and found an error. now we have the same result.
icheckedmyresultagainandfoundanerror.nowwehavethesameresult.

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